What is the extreme f(x)=Y(x)=x^2-5x+6 ?

The extreme f(x)=Y(x)=x^2-5x+6 is Minimum(5/2 ,-1/4)

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extreme f(x)=Y(x)=x^2-5x+6

extreme

Minimum (\frac{5}{2}, - \frac{1}{4})

Solution steps

f^{'} (x) = 2 x - 5

Find intervals:

Decreasing

: - \infty < x < \frac{5}{2},

Increasing

: \frac{5}{2} < x < \infty

Plug

x = \frac{5}{2}

into

x^{2} - 5 x + 6 : - \frac{1}{4}

Minimum (\frac{5}{2}, - \frac{1}{4})

What is the extreme f(x)=Y(x)=x^2-5x+6 ?
The extreme f(x)=Y(x)=x^2-5x+6 is Minimum(5/2 ,-1/4)